Here is a trivial example: Assume that $X$ is a finite set, $\mathcal{A} = P(X)$ and $\mu$ is the counting measure. We can give $P(X)$ the structure of an undirected graph: Join $A$ and $B$ by an edge if they differ by a point. Then the metric induced on $\mathcal{A}$ by the measure is exactly the graph metric (i.e the distance between two points is the length of the shortest path).

This is just the metric induced by the norm of the Banach space $L^1(X)$ on the closed (thus complete) metric subspace of {0,1}-valued functions (i.e. characteristic functions of measurable sets).
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Pietro MajerNov 4 '11 at 10:13

4 Answers
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To elaborate on Pietro's comment: if $Y$ is a set, and $S$ is a subset, define the cut metric $d_S$ as the pseudo-metric given by $d_S(y,y')=|1_S(y)-1_S(y')|$, where $1_S$ is the characteristic function of $S$. Let $Cut(Y)$ be the cone generated by cut metrics in the space of non-negative, symmetric kernels on $Y$. If $Y$ is finite (think of $Y$ as your $P(X)$), then a pseudo-metric $d$ on $y$ belongs to $Cut(Y)$ iff there is a map $f:Y\rightarrow L^1$ such that $d(y,y')=\|f_y-f_{y'}\|_1$. See M. Deza and M. Laurent, "Geometry of cuts and metrics", Springer 1997.

Actually we are not very far from the notion of measured wall space, i.e. a Borel measure $\nu$ on the space $P(X)$ (where $X$ is now an arbitrary set) such that, for every $x,x'\in X$, we have $\nu(E_x\Delta E_{x'})<+\infty$, where $E_x$ is the set of subsets through $x$. Then $d_W(x,x')=:\nu(E_x\Delta E_{x'})$ defines a pseudo-metric on $X$, called the wall metric. The relation between measured wall spaces, median metric spaces and embeddability into $L^1$ has been worked out in a nice paper by I. Chatterji, C. Drutu and F. Haglund, see http://fr.arxiv.org/PS_cache/arxiv/pdf/0704/0704.3749v4.pdf

This metric recovers the measure space up to measure-preserving transformations. Fix a point to be $0$. You can take unions and intersections relative to that point, using only the metric. For instance, $X\cap Y$ is the point farthest from $0$ such that two triangle inequalities are exact: $d(X,P)+d(P,0)=d(X,0)$ and $d(Y,P)+d(P,0)=d(Y,0)$. Similarly for anything else you want to do with sets. So you get the entire sigma-algebra structure, modulo sets of measure $0$.

However, that structure isn't very much. Any measure-preserving transformation between two measure spaces is an isomorphism from this perspective.

The result of taking the quotient of a sigma-algebra by the ideal of null sets is known as a measure algebra. And it is a rather rich structure. Fremlin devotes more than 600 pages (the entire 3rd volume) on measure algebras in his encyclopedic series on measure theory. How much structure remains is conclusively answered by Maharam's classification theorem: ncbi.nlm.nih.gov/pmc/articles/PMC1078424 (Theorems 1 and 2)
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Michael GreineckerNov 4 '11 at 20:39

Arguably, that should be the answer to the question.
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Will SawinNov 4 '11 at 21:57

Complementing the previous answer and comment: These metric is used in the context of shape optimization problems (in the case of $X$ a subset of $\mathbb{R}^d$ with the Lebesgue measure). In "Shapes and Geometries" by Delfour and Zolesio and you can find several interesting results in that book. E.g. the $L^p$-norm can be used ($1\leq p < \infty$) instead of $L^1$ in Pietro's comment and lead to the same topology, you can approximate any Lebesgue measurable set by $C^\infty$-domains in these metrics, the characteristic functions of convex sets form a closed subset.

The measure space can be infinite--you just need to restrict to the sets of finite measure.

Pietro pointed out that the resulting metric space embeds isometrically into $L_1(\mu)$. Conversely, $L_1(\mu)$ embeds isometrically into the sets of finite measure in the measure space $(X\times \Bbb{R}, \mathcal{A}, \mu \times m)$ ($m$ is Lebesgue measure)--map the function $f$ to the region under its graph.
As Alain points out, you can then snowflake into $L_2$. This is an easy way of seeing that $L_1$ with the snowflake distance $d(f,g) := \|f-\|g_2^{1/2}$ embeds isometrically into $L_2$, and hence $L_1$ uniformly embeds into $L_2$.